54. 螺旋矩阵 @Python @Go

原题

https://leetcode.cn/problems/spiral-matrix/description/

思路

循环

复杂度

时间：O(n)
空间：O(n)

Python代码

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        ans = []
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        # 方向
        d = 0
        m = len(matrix)
        n = len(matrix[0])
        count = 0
        i, j = 0, 0
        while count < m * n:
            ans.append(matrix[i][j])
            # 标记已走的
            matrix[i][j] = 1000
            count += 1
            if (
                i + directions[d][0] < 0
                or i + directions[d][0] == m
                or j + directions[d][1] < 0
                or j + directions[d][1] == n
            ):
                d = (d + 1) % 4
            elif matrix[i + directions[d][0]][j + directions[d][1]] == 1000:
                d = (d + 1) % 4
            i += directions[d][0]
            j += directions[d][1]
        return ans

Go代码

func spiralOrder(matrix [][]int) []int {
	var ans []int
	directions := [][]int{{0, 1}, {1, 0}, {0, -1}, {-1, 0}}
	// 方向
	d := 0
	m, n := len(matrix), len(matrix[0])
	count := 0
	i, j := 0, 0
	for count < m*n {
		ans = append(ans, matrix[i][j])
		// 标记已走过的
		matrix[i][j] = 1000
		count++
		if i+directions[d][0] < 0 || i+directions[d][0] == m ||
			j+directions[d][1] < 0 || j+directions[d][1] == n ||
			matrix[i+directions[d][0]][j+directions[d][1]] == 1000 {
			d = (d + 1) % 4
		}
		i += directions[d][0]
		j += directions[d][1]
	}
	return ans
}